Ta có: \(\left(\sin\alpha+\cos\alpha\right)^2=\sin^2\alpha+\cos^2\alpha+2\sin\alpha.\cos\alpha\)\(=1+2.\frac{1}{2}=1+1=2\)

=> \(\sin\alpha+\cos\alpha=\sqrt{2}\)=> \(\sin\alpha=\sqrt{2}-\cos\alpha\)

=> \(\sin\alpha.\cos\alpha=\left(\sqrt{2}-\cos\alpha\right).\cos\alpha=\sqrt{2}.\cos\alpha-\cos^2\alpha=\frac{1}{2}\)

=> \(\cos^2\alpha-\sqrt{2}\cos\alpha+\frac{1}{2}=0\)

Xong bạn giải phương trình bậc 2 => \(\cos\alpha=\frac{\sqrt{2}}{2}\)=> \(\alpha=45^o\)

\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\)

\(\cot=\frac{1}{\tan}=\frac{1}{\frac{2\sqrt{5}}{5}}=\frac{\sqrt{5}}{2}\)

Đề sai nhé, phải là 16,3.

\(7\sin\alpha+13\cos\left(90-\alpha\right)=16,3\)

\(\Leftrightarrow7\sin\alpha+13\sin\alpha=16,3\)

\(\Leftrightarrow20\sin\alpha=16,3\)

\(\Leftrightarrow\sin\alpha=0,815\)

\(\Rightarrow\alpha\approx55\left(độ\right)\)

Viết sai rồi.

sin [a(90-a)]

Đề lỗi font. Bạn cần chỉnh sửa lại bằng công thức toán để được hỗ trợ tốt hơn.

Đề bị lỗi rồi bạn ơi

Ko biết làm

Bài 1: 

\(\cos\alpha=\dfrac{4}{5}\)

\(\tan\alpha=\dfrac{3}{4}\)

\(\cot\alpha=\dfrac{4}{3}\)

1.Ta có :

\(\cot41=\tan49\) ; \(\cot46=\tan44\)

sắp xếp :\(\tan27< \tan44< \tan47< \tan49\)\(\Rightarrow\tan27< \cot46< \tan47< \cot41\)

2.ta có 

\(\cos28=\sin62;\cos41=\sin49\)

\(A=\cos^228+\cos^241+\cos^262+\cos^249\)

\(\Rightarrow A=\sin^262+\cos^262+\sin^249+\cos^249\)

\(\Rightarrow A=1+1=2\)

a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)